YES

Problem:
 c() -> f()
 f() -> g()

Proof:
 DP Processor:
  DPs:
   c#() -> f#()
  TRS:
   c() -> f()
   f() -> g()
  TDG Processor:
   DPs:
    c#() -> f#()
   TRS:
    c() -> f()
    f() -> g()
   graph:
    
   Qed